The theorem states
If a continuous map $f\,:\,A\to\Bbb R$ with $A$ a closed subset of the
normal topological space $X$ carries the standard topology, some
continuous extension $F\,:\,X\to\Bbb R$ of $f$ satisfies $\sup\{|f(a)|\mid a\in A\}=\sup\{|F(X)|\mid x\in X\}$.
The lemma states
If for subsets $A,\,B$ of a topological space $X$ some continuous Urysohn function $F\,:\,X\to[0,\,1]$ has $\forall (a,\,b)\in A\times B((f(a),\,f(b))=(0,\,1))$, the same is true of the closures of $A,\,B$; in particular there exist respective neighbourhoods $U,\,V$ of $A,\,B$ with $U\cap V=\emptyset$.
I came across this problem in Munkres Topology second edition. I am having a hard time with this problem and would like some help on it. Would just writing out the proof for Urysohn's Lemma be enough to solve this problem? Or does it require something else?
Best Answer
It's clear that the Tietze extension theorem implies Urysohn's lemma: if $A$ and $B$ are disjoint closed sets of a normal space $X$, define $f: A \cup B \to \mathbb{R}$ by $$f(x)=\begin{cases}0 & x \in A\\ 1 & x \in B\end{cases}$$ and note that $A\cup B$ is closed in $X$ and $f$ is continuous (the gluing lemma for closed sets implies this) and so it has a continuous extension $F: X \to \mathbb{R}$ (forgetting the $\sup$-property, that we don't need) and this extension is a Urysohn function for $A$ and $B$ by definition.